Linear Algebra Examples

$5 x - 4 y = - 3$ , $8 x - y = 2$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 5 & - 4 & - 3 \\ 8 & - 1 & 2 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{- 4}{5} & \frac{- 3}{5} \\ 8 & - 1 & 2 \end{array}]$

Step 2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 8 & - 1 & 2 \end{array}]$

Step 2.2

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 2.2.1

Perform the row operation $R_{2} = R_{2} - 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 8 - 8 \cdot 1 & - 1 - 8 (- \frac{4}{5}) & 2 - 8 (- \frac{3}{5}) \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & \frac{27}{5} & \frac{34}{5} \end{array}]$

Step 2.3

Multiply each element of $R_{2}$ by $\frac{5}{27}$ to make the entry at $2, 2$ a $1$ .

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Step 2.3.1

Multiply each element of $R_{2}$ by $\frac{5}{27}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ \frac{5}{27} \cdot 0 & \frac{5}{27} \cdot \frac{27}{5} & \frac{5}{27} \cdot \frac{34}{5} \end{array}]$

Step 2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{4}{5} & - \frac{3}{5} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 2.4

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 2.4.1

Perform the row operation $R_{1} = R_{1} + \frac{4}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{4}{5} \cdot 0 & - \frac{4}{5} + \frac{4}{5} \cdot 1 & - \frac{3}{5} + \frac{4}{5} \cdot \frac{34}{27} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 2.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{11}{27} \\ 0 & 1 & \frac{34}{27} \end{array}]$

Step 3

Use the result matrix to declare the final solution to the system of equations.

$x = \frac{11}{27}$

$y = \frac{34}{27}$

Step 4

The solution is the set of ordered pairs that make the system true.

$(\frac{11}{27}, \frac{34}{27})$

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